Proof of $\lim_{n\to\infty}\frac{e^nn!}{n^n\sqrt{n}}=\sqrt{2\pi}$ I'm looking for a proof of the following limit:
$$\lim_{n\to\infty}\frac{e^nn!}{n^n\sqrt{n}}=\sqrt{2\pi}$$
This follows from Stirling's Formula, but how can it be proven?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With the
  Stirling's Approximation for $n!$ $\ds{\approx \root{2\pi}n^{n + 1/2}\expo{-n}}$

\begin{align}
\color{#66f}{\large\lim_{n\ \to\ \infty}{\expo{n}n! \over n^{n}\root{n}}}
=\lim_{n\ \to\ \infty}
{\expo{n}\color{#c00000}{\root{2\pi}n^{n + 1/2}\expo{-n}} \over n^{n + 1/2}}
=\color{#66f}{\large\root{2\pi}} \approx {\tt 2.5066}
\end{align}
